Find a number field whose unit group is isomorphic to $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}.$

I'm trying to use Dirichlet's Unit Theorem to solve this problem. It states that if $K$ is a number field of signature $(r,s)$ and $\mu_K$ is the set of roots of unity in $K$, then the unit group $\mathcal{O}_K^{\times}$ of the ring of integers is isomorphic to $\mu_K \times \mathbb{Z}^{r+s−1}$ as an abelian group. So I suppose I want $r+s-1=1,$ or $r+s=2$. This forces $(r,s)=(0,2)$ because if there's at least one real embedding then $\mu_K$ is just $\{\pm 1\}$ so not $\mathbb{Z}/4\mathbb{Z}$.

Therefore I need a number field of degree $4$ with four complex embeddings and whose set of roots of unity is isomorphic to $\mathbb{Z}/4\mathbb{Z}$. The cyclotomic field $\mathbb{Q}(\zeta_5)$ doesn't work because it has more than $4$ elements in its set of roots of unity (and $\mathbb{Q}(\zeta_4)$ doesn't work because it doesn't have degree $4$). I suppose it will have to be $\mathbb{Q}(\alpha)$ where the minimal polynomial of $\alpha$ has degree $4$ but I haven't been able to find an example. Hints would be appreciated.

  • 3
    $\begingroup$ If the set of roots of unity in your $K$ has to be cyclic of order $4$ then it has to be $\{\pm1,\pm i\}$ under any complex embedding. Thus, I would search your $K$ among those of the form $\Bbb Q(i,\sqrt{d})$ with $d<0$ to start with. $\endgroup$ – Andrea Mori Mar 15 at 20:35
  • $\begingroup$ So perhaps I could take $\mathbb{Q}(i, \sqrt{2})$? That has degree $4$ and four complex embeddings. Its set of roots of unity includes $\pm 1,\pm i$, so it would suffice to show there are no other roots of unity in this set. Any other root of unity would need to be a primitive $m$th root of unity with $m=3$ or $m \geq 5$. I'm not sure how to show this is impossible though. $\endgroup$ – AlephNull Mar 15 at 21:12
  • $\begingroup$ Does $\Bbb Q(\zeta_8)$ get you where you want to be? $\endgroup$ – Robert Shore Mar 15 at 22:21
  • $\begingroup$ @RobertShore I considered that but I'm not sure what made me refute it. It certainly has degree $4$ and four complex embeddings. But doesn't that have more than $4$ roots of unity? $\endgroup$ – AlephNull Mar 15 at 22:50
  • $\begingroup$ @AlephNull I think your example of $\mathbb{Q}(i, \sqrt{2})$ works. You can check that the roots of unity in $\mathbb{Q}(\zeta_n)$ are $\mu_n$ if $n$ is even and $\mu_{2n}$ if $n$ is odd. (Intuitively, this is because multiplying by $-1$ will double the order of an odd root of unity) $\endgroup$ – vxnture Mar 15 at 23:00

You are almost there! Since $K$ has a torsion element of order $4$, it contains $\zeta_4$ and thus contains $\mathbf{Q}(\zeta_4)$. Dirichlet's unit theorem then says that $K$ must be degree $4$ and thus a quadratic extension of $\mathbf{Q}(\zeta_4)$.

Now suppose that $K$ is any quadratic extension of $\mathbf{Q}(\zeta_4) = \mathbf{Q}(\sqrt{-1})$. The signature of $K$ is $(0,2)$ and so $K$ has unit rank one. Also, $K$ has an element $\zeta_4$ of order $4$. The only thing that remains is to find a $K$ that doesn't have any extra torsion. But the torsion subgroup of a number field is always cyclic and generated by a $m$th root of unity, or a $4n$th root of unity in this case since we already have a $4$th root of unity. So you just have to ensure that

$$\mathbf{Q}(\zeta_{4n}) \not\subset K$$

for any $n > 1$. The degree of $\mathbf{Q}(\zeta_{4n})$ is (Euler's $\varphi$ function) $\varphi(4n)$. This is $> 4$ for $n \ge 4$. So the answer is:

$K$ can be any quadratic extension of $\mathbf{Q}(\zeta_4)$ which doesn't equal $\mathbf{Q}(\zeta_8)$ or $\mathbf{Q}(\zeta_{12})$.

Since $\mathbf{Q}(\zeta_8) = \mathbf{Q}(\sqrt{-1},\sqrt{2}) = \mathbf{Q}(\sqrt{-1},\sqrt{-2})$ and $\mathbf{Q}(\zeta_{12}) = \mathbf{Q}(\sqrt{-1},\sqrt{-3}) = \mathbf{Q}(\sqrt{-1},\sqrt{2})$, you can find many such $K$, for example $K = \mathbf{Q}(\sqrt{-1},\sqrt{d})$ for any squarefree $\pm d > 3$. These are not the only examples --- the others are precisely all the quadratic extensions $K/\mathbf{Q}(\sqrt{-1})$ which are non-Galois over $\mathbf{Q}$ such as $\mathbf{Q}(i,\sqrt{3 + 4 i})$.

  • $\begingroup$ Perfect answer, and this only uses results proved in my course. $\endgroup$ – AlephNull Mar 18 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.